Derivative Review
Y
which reads rate of
X
change of Y with respect to X. Since constant cannot change, then the derivative of a
Y
 0  n(bx n 1 ) . For instance,
constant is zero. In general, if y  a  bx n ,
X
2
21
y  5  3x  0  2(3x )  6 x or y  5a  3ax 6  0  18ax 5 .
Derivative shows the rate of change and it is represented as
Note that we also use f’(x) to mean derivative of f(x) with respect to x.
Rule #1:
Derivative of f x.g x  f ' xg x  g ' x f x
Rule #2:
If hx  
f x 
, then
g x 
derivative of h(x) with respect to x is:
h' ( x ) 
f ' ( x) g ( x)  g ' ( x) f ( x)
g ( x) 2
Rule #3:
If y  f (u ) and u  g (x) the derivative of y with respect to x is:
y' 
y u
*
u  x
Example #1:


Find the derivative of h  2 x 3  4 x 2 3x 5  x 2
Using Rule #1:



h  f x  * g x ; f x   2 x 3  4 x 2 ,

g  x   3x 5  x 2

and h'  f ' xg x  g ' x *  f x


 

h'  6 x 2  8x 3x 5  x 2  15 x 4  2 x 2 x 3  4 x 2

Example #2:
Find the derivative of h 
2x3  4
x 2  4x  1
Using Rule #2,
6x x
h' 
2
2


 4 x  1  2 x  4 2 x 3  4
x
2

 4x  1

2
Example #3:


5
Find the derivative of y  2 x 3  5 x 2  4 .

Using Rule #3, let’s define u  2 x 3  5 x 2  4
Then y  u 5
y y u

*
x u x
 5 u 4 6 x 2  10 x
 
 52 x  5 x
3
2

 4  6 x
4
2
 10 x


Example #4:
Find derivative of y 
1
4 x  5x 2  7 x  8
3
y  4 x 3  5 x 2  7 x  8 , let u  4 x 3  5 x 2  7 x  8


1
y y u

*
  u  2 12 x 2  10 x  7
x u x
 
 4 x  5 x  7 x  8 12 x
 12 x  10 x  7 

4x  5x  7 x  8
3
2
2
2

 10 x  7

2
3
2
2
Example #5:
 2x  1 
Find the derivative of y  
 .
 3x  1 
4
First define u 
2x  1
or y  u 4
3x  1
 23x  1  32 x  1
y y u

*
 4u 3 * 

x u x
3x  12


 6x  2  6x  3 


 3x  12



3
 202 x  1
y' 
3x  15
 2x  1 
 4

 3x  1 
3